开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)｜3.2.5(安卓)APP下载

返回

zoewong · 2022年04月04日

為什麼不用sample mean 15% - 10% /25%/10

  1. CFA I
  2. Quantitative

NO.PZ2015120604000130

问题如下：

We want to use z-statistic to construct confidence interval for a normally distribution. Assume the sample size is 100, sample mean is 15% and the standard deviation of sample is 25%. The significance level is supposed to be 10% , the confidence interval is  :

选项：

A.

- 10.88% to 19.13%.

B.

10.88% to 19.13%.

C.

10.88% to 20.57%.

解释：

B is correct.

In this case, we can know that the Confidence Interval=[Point Estimate +/- (reliability factor)*Standard error]=

x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13%

為什麼不用sample mean 15% - 10% /25%/10

添加评论
1 个答案
已采纳答案

星星_品职助教 · 2022年04月04日

同学你好，

正确的置信区间公式如下。逐项代入题干中数据即可。

正态分布下，significance level为10%所对应的reliability factor（即critical value） 为1.65.



  • 1

    回答

  • 0

    关注

  • 326

    浏览
相关问题

NO.PZ2015120604000130问题如下We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis :A.- 10.88% to 19.13%.B.10.88% to 19.13%.C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]= 什么时候会考这个区间，一般问这个区间的问题是怎么问的这个区间的标准差指的是总体标准差吗

2023-01-18 11:01 1 · 回答

NO.PZ2015120604000130 问题如下 We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis : A.- 10.88% to 19.13%. B.10.88% to 19.13%. C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]= SignificanLevel都是要除以2吗？

2022-11-01 18:08 1 · 回答

NO.PZ2015120604000130 问题如下 We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis : A.- 10.88% to 19.13%. B.10.88% to 19.13%. C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]=x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13% 请题干以及该题涉及的原理所对应知识点

2022-10-13 23:10 1 · 回答

NO.PZ2015120604000130 问题如下 We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis : A.- 10.88% to 19.13%. B.10.88% to 19.13%. C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]=x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13% 这里的Stanrerror 不应该是25%/ √100 吗？解析里的式子是什么意思

2022-10-09 10:47 1 · 回答

NO.PZ2015120604000130问题如下We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis :A.- 10.88% to 19.13%.B.10.88% to 19.13%.C.10.88% to 20.57%.B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]=x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13%老师，既然题目说是正态分布，而且是选择题。想问下能否利用正态分布的特性，代入解题？正太分布中轴是均值，题干给出的置信区间应该也是轴对称。直接带入，分位点-均值19.13-15=4.13均值-分位点到均值的距离15-4.13=10.87然后选出最符合的的

2022-09-27 15:46 1 · 回答